Exercise 1946Geometry › Plane geometryL1Added 4 Nov 2026

The Euler line, by vectors

#Euler line#orthocentre#centroid#circumcentre#vectors

Statement

Let $O$ , $G$ , $H$ be the circumcentre, centroid and orthocentre of a triangle $ABC$ . Prove that $O,G,H$ are collinear with $\overrightarrow{OH}=3\overrightarrow{OG}$ . Deduce that $AH=2\cdot OM$ , where $M$ is the midpoint of $BC$ .

Hint

Take $O$ as the origin and denote by $\mathbf a,\mathbf b,\mathbf c$ the position vectors of $A,B,C$ , which all have the same length $R$ . Show that the point with vector $\mathbf a+\mathbf b+\mathbf c$ lies on all three altitudes.

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Solution

Idea. With the circumcentre as origin, the orthocentre has the very simple vector $\mathbf h=\mathbf a+\mathbf b+\mathbf c$ .

The orthocentre. Let $\mathbf h=\mathbf a+\mathbf b+\mathbf c$ . Then
$(\mathbf h-\mathbf a)\cdot(\mathbf b-\mathbf c)=(\mathbf b+\mathbf c)\cdot(\mathbf b-\mathbf c)=|\mathbf b|^2-|\mathbf c|^2=R^2-R^2=0.$
So $H'$ (the point with vector $\mathbf h$ ) lies on the altitude from $A$ . By symmetry it lies on the altitudes from $B$ and $C$ . Hence $H'=H$ .

The centroid. $\mathbf g=\frac{\mathbf a+\mathbf b+\mathbf c}3=\frac{\mathbf h}3$ , i.e. $\overrightarrow{OH}=3\overrightarrow{OG}$ . So $O,G,H$ are collinear and $G$ divides $OH$ in the ratio $1:2$ .

The second statement. $\overrightarrow{AH}=\mathbf h-\mathbf a=\mathbf b+\mathbf c=2\overrightarrow{OM}$ , hence $AH=2\,OM$ and $AH\parallel OM$ . $\blacksquare$

Remark. The line through $O,G,H$ is the Euler line. The relation $AH=2R|\cos A|$ also follows.

Lesson. Choose the origin at the circumcentre: the orthocentre is then the sum of the vertices.

The Euler line, by vectors

Statement

Solution

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